Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 TRUE
  ↳25 QDP

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(f(X, g(X), Y)) → F(Y, Y, Y)
ACTIVE(g(X)) → G(active(X))
ACTIVE(g(X)) → ACTIVE(X)
G(mark(X)) → G(X)
PROPER(f(X1, X2, X3)) → F(proper(X1), proper(X2), proper(X3))
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(g(X)) → G(proper(X))
PROPER(g(X)) → PROPER(X)
F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
G(ok(X)) → G(X)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 5 SCCs with 6 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(ok(X1), ok(X2), ok(X3)) → F(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3)  =  F(x3)
ok(x1)  =  ok(x1)
active(x1)  =  active(x1)
f(x1, x2, x3)  =  f(x3)
g(x1)  =  x1
mark(x1)  =  mark(x1)
b  =  b
c  =  c
proper(x1)  =  proper(x1)
top(x1)  =  top(x1)

Lexicographic Path Order [LPO].
Precedence:
F1 > top1
b > [ok1, active1, f1, mark1, proper1] > c > top1


The following usable rules [FROCOS05] were oriented:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(7) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

G(ok(X)) → G(X)
G(mark(X)) → G(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


G(ok(X)) → G(X)
G(mark(X)) → G(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
G(x1)  =  G(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  mark(x1)
active(x1)  =  active(x1)
f(x1, x2, x3)  =  f(x3)
g(x1)  =  x1
b  =  b
c  =  c
proper(x1)  =  proper(x1)
top(x1)  =  top(x1)

Lexicographic Path Order [LPO].
Precedence:
b > c > [ok1, mark1, active1, f1, proper1]
top1 > [ok1, mark1, active1, f1, proper1]


The following usable rules [FROCOS05] were oriented:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(12) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE

(15) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(g(X)) → PROPER(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(f(X1, X2, X3)) → PROPER(X2)
PROPER(f(X1, X2, X3)) → PROPER(X1)
PROPER(f(X1, X2, X3)) → PROPER(X3)
PROPER(g(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1)  =  PROPER(x1)
f(x1, x2, x3)  =  f(x1, x2, x3)
g(x1)  =  g(x1)
active(x1)  =  x1
mark(x1)  =  mark
b  =  b
c  =  c
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
proper1 > [g1, c] > [f3, mark, b]


The following usable rules [FROCOS05] were oriented:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(17) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(18) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(19) TRUE

(20) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ACTIVE(g(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(21) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(g(X)) → ACTIVE(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1)  =  ACTIVE(x1)
g(x1)  =  g(x1)
active(x1)  =  active(x1)
f(x1, x2, x3)  =  f(x2, x3)
mark(x1)  =  mark(x1)
b  =  b
c  =  c
proper(x1)  =  proper(x1)
ok(x1)  =  x1
top(x1)  =  top

Lexicographic Path Order [LPO].
Precedence:
active1 > [g1, f2, mark1, b, proper1] > c
active1 > [g1, f2, mark1, b, proper1] > top


The following usable rules [FROCOS05] were oriented:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

(22) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(24) TRUE

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(f(X, g(X), Y)) → mark(f(Y, Y, Y))
active(g(b)) → mark(c)
active(b) → mark(c)
active(g(X)) → g(active(X))
g(mark(X)) → mark(g(X))
proper(f(X1, X2, X3)) → f(proper(X1), proper(X2), proper(X3))
proper(g(X)) → g(proper(X))
proper(b) → ok(b)
proper(c) → ok(c)
f(ok(X1), ok(X2), ok(X3)) → ok(f(X1, X2, X3))
g(ok(X)) → ok(g(X))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.